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| | <span class="exam">(b) <math>\int \frac{\sqrt{x+1}}{x}~dx</math> | | <span class="exam">(b) <math>\int \frac{\sqrt{x+1}}{x}~dx</math> |
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
| + | <hr> |
| − | !Foundations:
| + | [[009B Sample Final 3, Problem 6 Solution|'''<u>Solution</u>''']] |
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| − | |Through partial fraction decomposition, we can write the fraction
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| − | |-
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| − | | <math style="vertical-align: -18px">\frac{1}{(x+1)(x+2)}=\frac{A}{x+1}+\frac{B}{x+2}</math>
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| − | |- | |
| − | |for some constants <math style="vertical-align: -4px">A,B.</math>
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| − | |}
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| − | '''Solution:''' | + | [[009B Sample Final 3, Problem 6 Detailed Solution|'''<u>Detailed Solution</u>''']] |
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| − | '''(a)'''
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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| − | !Step 1:
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| − | |-
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| − | |First, we factor the denominator to get
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| − | |-
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| − | | <math>\int \frac{3x-1}{2x^2-x}~dx=\int \frac{3x-1}{x(2x-1)}.</math>
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| − | |-
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| − | |We use the method of partial fraction decomposition.
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| − | |-
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| − | |We let
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| − | | <math>\frac{3x-1}{x(2x-1)}=\frac{A}{x}+\frac{B}{2x-1}.</math>
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| − | |-
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| − | |If we multiply both sides of this equation by <math style="vertical-align: -5px">x(2x-1),</math> we get
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| − | |-
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| − | | <math>3x-1=A(2x-1)+Bx.</math>
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| − | |}
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| − |
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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| − | !Step 2:
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| − | |-
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| − | |Now, if we let <math style="vertical-align: -4px">x=0,</math> we get <math style="vertical-align: -2px">A=1.</math>
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| − | |-
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| − | |If we let <math style="vertical-align: -13px">x=\frac{1}{2},</math> we get <math style="vertical-align: -2px">B=1.</math>
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| − | |Therefore,
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| − | | <math>\frac{3x-1}{x(2x-1)}=\frac{1}{x}+\frac{1}{2x-1}.</math>
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| − | |}
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| − |
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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| − | !Step 3:
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| − | |-
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| − | |Now, we have
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| − | |-
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| − | | <math>\begin{array}{rcl}
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| − | \displaystyle{\int \frac{3x-1}{2x^2-x}~dx} & = & \displaystyle{\int \frac{1}{x}+\frac{1}{2x-1}~dx}\\
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| − | &&\\
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| − | & = & \displaystyle{\int \frac{1}{x}~dx+\int \frac{1}{2x-1}~dx}\\
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| − | &&\\
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| − | & = & \displaystyle{\ln |x|+\int \frac{1}{2x-1}~dx.}
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| − | \end{array}</math>
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| − | |-
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| − | |Now, we use <math style="vertical-align: 0px">u</math>-substitution.
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| − | |Let <math style="vertical-align: -1px">u=2x-1.</math>
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| − | |-
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| − | |Then, <math style="vertical-align: -1px">du=2dx</math> and <math style="vertical-align: -13px">\frac{du}{2}=dx.</math>
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| − | |-
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| − | |Hence, we have
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| − | | <math>\begin{array}{rcl}
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| − | \displaystyle{\int \frac{3x-1}{2x^2-x}~dx} & = & \displaystyle{\ln |x|+\frac{1}{2}\int \frac{1}{u}~du}\\
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| − | &&\\
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| − | & = & \displaystyle{\ln |x|+\frac{1}{2}\ln |u|+C}\\
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| − | &&\\
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| − | & = & \displaystyle{\ln |x|+\frac{1}{2}\ln |2x-1|+C.}
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| − | \end{array}</math>
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| − | |}
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| − |
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| − | '''(b)'''
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| − |
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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| − | !Step 1:
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| − | |-
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| − | |We begin by using <math>u</math>-substitution.
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| − | |Let <math>u=\sqrt{x+1}.</math>
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| − | |Then, <math>u^2=x+1</math> and <math>x=u^2-1.</math>
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| − | |Also, we have
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| − | | <math>\begin{array}{rcl}
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| − | \displaystyle{du} & = & \displaystyle{\frac{1}{2} (x+1)^{\frac{-1}{2}}dx}\\
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| − | &&\\
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| − | & = & \displaystyle{\frac{1}{2\sqrt{x+1}}dx}\\
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| − | &&\\
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| − | & = & \displaystyle{\frac{1}{2u}dx.}
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| − | \end{array}</math>
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| − | |-
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| − | |Hence, <math>dx=2udu.</math>
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| − | |-
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| − | |Using all this information, we get
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| − | | <math>\int \frac{\sqrt{x+1}}{x}~dx=\int \frac{2u^2}{u^2-1}~du.</math>
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| − | |}
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| − |
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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| − | !Step 2:
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| − | |-
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| − | |Now, we have
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| − | |-
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| − | | <math>\begin{array}{rcl}
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| − | \displaystyle{\int \frac{\sqrt{x+1}}{x}~dx} & = & \displaystyle{\int \frac{2u^2-2+2}{u^2-1}~du}\\
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| − | &&\\
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| − | & = & \displaystyle{\int \frac{2(u^2-1)}{u^2-1}~du+\int \frac{2}{u^2-1}~du}\\
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| − | &&\\
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| − | & = & \displaystyle{\int 2~du+\int \frac{2}{u^2-1}~du}\\
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| − | &&\\
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| − | & = & \displaystyle{2u+\int \frac{2}{u^2-1}~du}\\
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| − | &&\\
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| − | & = & \displaystyle{2\sqrt{x+1}+\int \frac{2}{(u-1)(u+1)}~du.}
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| − | \end{array}</math>
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| − | |}
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| − |
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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| − | !Step 3:
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| − | |-
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| − | |Now, for the remaining integral, we use partial fraction decomposition.
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| − | |-
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| − | |Let <math>\frac{2}{(x-1)(x+1)}=\frac{A}{x+1}+\frac{B}{x-1}.</math>
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| − | |-
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| − | |Then, we multiply this equation by <math>(x-1)(x+1)</math> to get
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| − | |-
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| − | | <math>2=A(x-1)+B(x+1).</math>
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| − | |-
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| − | |If we let <math>x=1,</math> we get <math>B=1.</math>
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| − | |If we let <math>x=-1,</math> we get <math>A=-1.</math>
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| − | |Thus, we have <math>\frac{2}{(x-1)(x+1)}=\frac{-1}{x+1}+\frac{1}{x-1}.</math>
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| − | |-
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| − | |Using this equation, we have
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| − | |-
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| − | | <math>\begin{array}{rcl}
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| − | \displaystyle{\int \frac{\sqrt{x+1}}{x}~dx} & = & \displaystyle{2\sqrt{x+1}+\int \frac{-1}{(u+1)}+\frac{1}{u-1}~du}\\
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| − | &&\\
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| − | & = & \displaystyle{2\sqrt{x+1}+\int \frac{-1}{(u+1)}~du+\int \frac{1}{u-1}~du.}\\
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| − | \end{array}</math>
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| − | |}
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| − |
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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| − | !Step 4:
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| − | |-
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| − | |To complete this integral, we need to use <math>u</math>-substitution.
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| − | |-
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| − | |For the first integral, let <math>t=u+1.</math> Then, <math>dt=du.</math>
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| − | |-
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| − | |For the second integral, let <math>v=u-1.</math> Then, <math>dv=du.</math>
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| − | |Finally, we integrate to get
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| − | |-
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| − | | <math>\begin{array}{rcl}
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| − | \displaystyle{\int \frac{\sqrt{x+1}}{x}~dx} & = & \displaystyle{2\sqrt{x+1}+\int \frac{-1}{t}~dt+\int \frac{1}{v}~dv}\\
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| − | &&\\
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| − | & = & \displaystyle{2\sqrt{x+1}+\ln|t|+\ln|v|+C}\\
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| − | &&\\
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| − | & = & \displaystyle{2\sqrt{x+1}+\ln|u+1|+\ln|u-1|+C}\\
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| − | &&\\
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| − | & = & \displaystyle{2\sqrt{x+1}+\ln|\sqrt{x+1}+1|+\ln|\sqrt{x+1}-1|+C.}
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| − | \end{array}</math>
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| − | |}
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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| − | !Final Answer:
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| − | |-
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| − | | '''(a)''' <math>\ln |x|+\frac{1}{2}\ln |2x-1|+C</math>
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| − | |-
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| − | | '''(b)''' <math>2\sqrt{x+1}+\ln|\sqrt{x+1}+1|+\ln|\sqrt{x+1}-1|+C</math>
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| − | |}
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| | [[009B_Sample_Final_3|'''<u>Return to Sample Exam</u>''']] | | [[009B_Sample_Final_3|'''<u>Return to Sample Exam</u>''']] |
